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2 years ago

How many lines of reflectional symmetry does a isosceles trapezoid have?

Mathematics

1 answer:

zalisa [80]2 years ago

4 0

How many lines of reflectional symmetry does a isosceles trapezoid have? 1

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Sharon is decreasing the size of a diagram of a leaf that is 30 centimeters long by 10 centimeters wide. If the reduced diagram

ivanzaharov [21]

10 : 30
4: x [x is the length]
10/4=2.5
30/2.5=x=12
length of the diagram is 12cm

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3 years ago

Find the value of x. NEED ANSWER ASAP WILL MARK BRAINLIEST! L, M ∈ KN

noname [10]

Answer:

the value of x in this image is 38

Step-by-step explanation:

we know that the total angles of triangles are 180, so m is 180-84=96, and 180-134=46, when added we get 142, which means the last angle is 38

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2 years ago

Read 2 more answers

Help pls this is my first question on here and I need help with this pls. I will give you the brainliest and a thanks!! :)) Can

docker41 [41]

Answer:

1. 18.85in

4. 25.13m

Step-by-step explanation:

I got you

So to find the circumference of any circle we use the equation 2pi(r)

2(3.1415)(3)= 18.85in for the first one because radius is half of the diameter which is 6 so half of 6 is 3

For 4. 4m is actually the radius so just plug that into the equation 2pi(4)=25.13m

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2 years ago

How the transformations of the functions f(x)=15tanx and g(x)=tanx+15 differ?

Irina-Kira [14]

The transformation of f(x)=15tanx would be achieved by dilating y=tanx graph 15 times. This transformation will cause y'=15 y. This will increase the value of dilation become 15 times of the original

The transformation of g(x)=15tanx would be achieved by shifting y=tanx graph 15 right. This transformation will cause x'=x+15. This will change the value from tan x into tan(x+15)

3 0

2 years ago

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Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

anyanavicka [17]

Answer:

a)The second student arrive 7 minutes and 36 seconds sooner

b)They have to walk 598,51 m

Step-by-step explanation:

Speed= Distance/Time

We are looking for time now

Time = Distance/Speed

First student

Time = 780 m/0.90 m/s=866,66 seconds

Second student

Time = 780 m/1.90 m/s=410,52 seconds

Difference=866,66-410,52= 456,14 seconds

In minutes

Difference= 456,14 seconds /60=7 minutes 36 seconds

Distance=Speed x Time

Speed¹ x Time¹ =Speed² x Time²

Time¹= Time² + 5.50 min

5.50 min =350 seconds

Time² =Speed¹ x Time¹ /Speed²

Time² =Speed¹ x (Time² + 350 seconds ) /Speed²

Time² =0.9 x (Time² + 350 seconds ) /1.9

Time² =0,4737 x (Time² + 350 seconds )

Time² =0,4737 Time² + 165,79 seconds

Time² - 0,4737 Time² = 165,79 seconds

0,5263 Time² = 165,79 seconds

Time² = 165,79 seconds/0,5263 =315,01 seconds

Distance=1,9 x 315,01 seconds = 598,51 m

5 0

2 years ago